如图（a）所示，在足够长的光滑水平面上，放置一长为L＝1m、质量为m₁＝0.5kg的木板A，一质量为m₂＝1kg的小物体B以初速度v₀滑上A的上表面的同时对A施加一个水平向右的力F，A与B之间的动摩擦因数为μ＝0.2，g＝10m/s²；小物体B在A上运动的路程S与F力的关系如图（b）所示。求：v₀、F₁、F₂。

（1）4m/s；（2）3N；（3）9N

（1）由图象可看出当F≤1N时，B物体在A板上的路程始终等于板长L，当F＝1N时，刚好不从A板右端掉下，此后A和B一起相对静止并加速运动。
设B物体的加速度为a₂，A板的加速度为a₁，分别由牛顿第二定律：
μm₂g＝m₂a₂ ············································································································ ①
F＋μm₂g＝m₁a₁ ······································································································· ②
设B运动的位移为S₂，A运动的位移为S₁，经过t时间两者速度均为v，根据运动学公式：
S_B＝t ············································································································· ③
S_A＝t ·················································································································· ④
v＝v₀－a₂t＝a₁t········································································································· ⑤
B在A上相对A向右运动的路程S＝S_B－S_A ······································································ ⑥
联立①②③④⑤⑥解得：S＝··············································································· ⑦
将F＝1N，S＝1m代入，解得：v₀＝4m/s
（2）根据⑦式分析可知，当1N≤F≤F₁时，随着F力增大，S减小，当F＝F₁时，出现S突变，说明此时A、B在达到共同速度后，恰好再次发生相对运动，B将会从A板左端掉下。
对A、B恰好发生相对运动时，B的加速度为a₂，则整体加速度也为a₂，由牛顿第二定律：
F₁＝(m₁＋m₂)a₂········································································································ ⑧
联立①⑧解得解得F₁＝3N
（3）此时B在A上运动的路程为S₁＝＝m
当F≥F₁时，物体B在A板上的路程为B相对A向右运动的路程的两倍。
故当F＝F₂时，将S＝0.5S₁代入⑦式解得：F₂＝9N
【评析】本题考查牛顿运动定律。滑块问题是物理模型中非常重要的模型，是学生物理建模能力培养的典型模型。滑块问题的解决非常灵活，针对受力分析、运动分析以及牛顿第二定律的掌握，还有相对运动的分析，特别是摩擦力的变化与转型，都是难点所在。本题通过非常规的图象来分析滑块的运动，能从图中读懂物体的运动。